typedef a function pointer type

Question

I want to declare a pointer type which point to a function, so I try:

typedef void (*print)(void); works perfect

void (*print)(void); p is a ponter variable , not a type.

typedef (void) (*print)(void); error expected identifier or ‘(’ before ‘void’

typedef void (*)(void) Print;

error: expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘_ attribute _’ before ‘Print’ .

My question is:

All your questions can pretty much be answered with "because the language designers decided it that way" — Per Johansson, Sep 05 '13 at 14:26
In an answer to another question, I tried to explain how to declare a function pointer. See http://stackoverflow.com/a/6905987/396551 — Sjoerd, Sep 05 '13 at 14:46

Juraj Blaho · Answer 1 · 2013-09-05T14:31:30.280

10

The correct way is:

typedef void (*print_function_ptr)(void)

and its usage for variable/parameter declaration is:

print_function_ptr p;

You don't need a typedef to declare a variable. You can directly write void (*p)(void) to declare a variable p pointing to a function taking void and returning void. However to declare a type alias / name for a pointer to function, typedef is the tool.
It does not mean anything it is not a valid C syntax.
Because it is not how C works. Typedefs in C mimics how variables are declared or defined.

edited Sep 05 '13 at 14:31

answered Sep 05 '13 at 14:25

Juraj Blaho

I hope declare a function pointer type , not a variable, is there a way to do this without typedef? – Lidong Guo Sep 05 '13 at 14:27
@LidongGuo: what do you mean by 'declare a function pointer type'? You can declare an object that is of type 'pointer to function' without using a `typedef`. There is no way other than using `typedef` to create a type name that you can use subsequently. – Jonathan Leffler Sep 05 '13 at 14:29
I hope to declare a type like a struct name, then I can use it declare lots of function pointer' – Lidong Guo Sep 05 '13 at 14:38
1

@LidongGuo then this answer shows you how to do that. After the typedef, you can just do this to have 5 function pointers with the same signature: `print_function_ptr p1, p2, p3, p4, p5;` – blubberdiblub Sep 24 '16 at 17:39

Jonathan Leffler · Answer 2 · 2013-09-05T14:34:15.190

No, you don't have to use a typedef to create an object of type 'pointer to function':
```
void somefunc(void (*pointer)(void))
{
    (*pointer)();
    pointer();
}
```
However, there is no way to create a name for a type other than by using a typedef. I suppose you could indulge in macro hackery to generate a 'pointer to function', but after the macro is expanded, you'd have a 'pointer to function' written out:
```
#define PTR_FUNC(func) void (*func)(void)
void somefunc(PTR_FUNC(pointer)) { ... }
```
The (void) notation as the type name is wrong. You don''t write: (int) x; and expect to declare a variable x -— it is a type cast. Same with the notation you're using in your typedef.
You can't write typedef void (*)(void) Print; because it is not an allowed C syntax. You also can't write typedef [32] int name; either — it isn't valid C syntax.

2 Answers2